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In$\triangle ABC$, $\angle C$ is a right angle. Prove that $r = \frac{a+b-c}{2}\;\text{and}\; r_c = \frac{a+b+c}{2}.$ Legs are named in traditional way.

My Work
As $r = \frac{A}{s}$. So, here $r = \frac{ab}{a+b+c}$. Is there any way to prove this is equal to $\frac{a+b-c}{2}$?. Or else how to solve this is also the second part?

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    Hint: $2ab=(a+b)^2-c^2$ by Pythagora's. Or geometrically, drop perpendiculars from the incenter onto the legs, and calculate the side of the small square that forms in terms of $a,b,c$. – 2017-01-31
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    What is $r_c$? $\text{}$ – 2017-01-31
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    @Displayname The $r$ of the excircle tangent to side $c$ – 2017-01-31

2 Answers 2

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The first part:

$$(a+b+c)(a+b-c)=(a+b)^2-c^2 = a^2+b^2 + 2ab -c^2=2ab$$ by Pythagora's theorem.

The second part: $r_c=\frac{A}{s-c}=\frac{ab}{a+b-c}$ and it works with absolutely the same calculation above.

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    I had done till this $a^2 + 2ab+b^2 - c^2$.. But Forgot about Pythagoras :P :P – 2017-01-31
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    Shouldn't it be $\frac{A}{s-c}$? – 2017-01-31
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    Yes, corrected. Thanks! – 2017-01-31
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$$r=\frac{2S}{a+b+c}=\frac{ab}{a+b+c}=\frac{ab(a+b-c)}{a^2+2ab+b^2-c^2}=\frac{a+b-c}{2}$$ $$r_c=\frac{2S}{a+b-c}=\frac{ab}{a+b-c}=\frac{ab(a+b+c)}{a^2+2ab+b^2-c^2}=\frac{a+b+c}{2}$$